I am facing a problem in MongoDB. Suppose, I have the following collection.
{ id: 1, issueDate: "07/05/2021", code: "31" },
{ id: 2, issueDate: "12/11/2020", code: "14" },
{ id: 3, issueDate: "02/11/2021", code: "98" },
{ id: 4, issueDate: "01/02/2021", code: "14" },
{ id: 5, issueDate: "06/23/2020", code: "14" },
{ id: 6, issueDate: "07/01/2020", code: "31" },
{ id: 7, issueDate: "07/05/2022", code: "14" },
{ id: 8, issueDate: "07/02/2022", code: "20" },
{ id: 9, issueDate: "07/02/2022", code: "14" }
The date field is in the format MM/DD/YYYY. My goal is to get the count of items with each season (spring (March-May), summer (June-August), autumn (September-November) and winter (December-February).
The result I'm expecting is:
count of fields for each season:
{ "_id" : "Summer", "count" : 6 } { "_id" : "Winter", "count" : 3 }
top 2 codes (first and second most recurring) per season:
{ "_id" : "Summer", "codes" : {14, 31} } { "_id" : "Winter", "codes" : {14, 98} }
How can this be done?
CodePudding user response:
You should never store date/time values as string, store always proper Date
objects.
You can use $setWindowFields opedrator for that:
db.collection.aggregate([
// Convert string into Date
{ $set: { issueDate: { $dateFromString: { dateString: "$issueDate", format: "%m/%d/%Y" } } } },
// Determine the season (0..3)
{
$set: {
season: { $mod: [{ $toInt: { $divide: [{ $add: [{ $subtract: [{ $month: "$issueDate" }, 1] }, 1] }, 3] } }, 4] }
}
},
// Count codes per season
{
$group: {
_id: { season: "$season", code: "$code" },
count: { $count: {} },
}
},
// Rank occurrence of codes per season
{
$setWindowFields: {
partitionBy: "$_id.season",
sortBy: { count: -1 },
output: {
rank: { $denseRank: {} },
count: { $sum: "$count" }
}
}
},
// Get only top 2 ranks
{ $match: { rank: { $lte: 2 } } },
// Final grouping
{
$group: {
_id: "$_id.season",
count: { $first: "$count" },
codes: { $push: "$_id.code" }
}
},
// Some cosmetic for output
{
$set: {
season: {
$switch: {
branches: [
{ case: { $eq: ["$_id", 0] }, then: 'Winter' },
{ case: { $eq: ["$_id", 1] }, then: 'Spring' },
{ case: { $eq: ["$_id", 2] }, then: 'Summer' },
{ case: { $eq: ["$_id", 3] }, then: 'Autumn' },
]
}
}
}
}
])
CodePudding user response:
I will give you clues,
- You need to use
$group
with_id
as$month
onissueDate
, use accumulator$sum
to get month wise count. - You can divide month by 3, to get modulo, using
$toInt, $divide
, then put them into category using$cond
.
CodePudding user response:
Another option:
db.collection.aggregate([
{
$addFields: {
"season": {
$switch: {
branches: [
{
case: {
$in: [
{
$substr: [
"$issueDate",
0,
2
]
},
[
"06",
"07",
"08"
]
]
},
then: "Summer"
},
{
case: {
$in: [
{
$substr: [
"$issueDate",
0,
2
]
},
[
"03",
"04",
"05"
]
]
},
then: "Spring"
},
{
case: {
$in: [
{
$substr: [
"$issueDate",
0,
2
]
},
[
"12",
"01",
"02"
]
]
},
then: "Winter"
}
],
default: "No date found."
}
}
}
},
{
$group: {
_id: {
s: "$season",
c: "$code"
},
cnt1: {
$sum: 1
}
}
},
{
$sort: {
cnt1: -1
}
},
{
$group: {
_id: "$_id.s",
codes: {
$push: "$_id.c"
},
cnt: {
$sum: "$cnt1"
}
}
},
{
$project: {
_id: 0,
season: "$_id",
count: "$cnt",
codes: {
"$slice": [
"$codes",
2
]
}
}
}
])
Explained:
- Add one more field for season based on $switch per month(extracted from issueDate string)
- Group to collect per season/code.
- $sort per code DESCENDING
- group per season to form an array with most recurring codes in descending order.
- Project the fields to the desired output and $slice the codes to limit only to the fist two most recurring.
Comment: Indeed keeping dates in string is not a good idea in general ...